A020473 -id:A020473 - cp's OEIS frontend

A038034 Number of compositions (ordered partitions) of 1 into {1/1, 1/2, 1/3, ..., 1/n}.

1, 2, 3, 7, 8, 52, 53, 288, 1209, 5247, 5248, 71395, 71396, 375779, 6957533, 52310862, 52310863, 1152622553, 1152622554, 45575902465, 1296407854551, 1580527987951, 1580527987952, 73245316681199, 584407520822198, 639887219617512, 11355804443049274, 516959218512416104, 516959218512416105, 29213061562205847736, 29213061562205847737, 886912328033731357358, 31286298736622399674197, 31349361777225437765677

Offset: 1

Christian G. Bower, Jun 15 1998

a(n) = number of Egyptian fractions 1 = 1/x_1 + ... + 1/x_k (for any k), with max{x_i}<=n.

			a(4) = 7 since there are seven compositions into parts {1/1, 1/2, 1/3, 1/4}:
1 = 1/1, 1 = 1/2 + 1/2, 1 = 1/3 + 1/3 + 1/3, 1 = 1/2 + 1/4 + 1/4, 1 = 1/4 + 1/2 + 1/4, 1 = 1/4 + 1/4 + 1/2, and 1 = 1/4 + 1/4 + 1/4 + 1/4.

Index entries for sequences related to Egyptian fractions

Cf. A002967, A020473, A092667, A092670.

a(n) = Sum_{i=1..n} A092667(i).

a(p) = a(p-1) + 1 for p prime. - Chai Wah Wu, Dec 27 2024

More terms from Max Alekseyev, Mar 02 2004

A037268 Sum of reciprocals of digits = 1.

Original entry on oeis.org

1, 22, 236, 244, 263, 326, 333, 362, 424, 442, 623, 632, 2488, 2666, 2848, 2884, 3366, 3446, 3464, 3636, 3644, 3663, 4288, 4346, 4364, 4436, 4444, 4463, 4634, 4643, 4828, 4882, 6266, 6336, 6344, 6363, 6434, 6443, 6626, 6633, 6662, 8248, 8284, 8428, 8482, 8824

Offset: 1

N. J. A. Sloane

This sequence has 1209 terms.

Intersection of A037264 and A034708: A214949(a(n))*A214950(a(n))*A168046(a(n)) = 1. - Reinhard Zumkeller, Aug 02 2012

Nathaniel Johnston, Table of n, a(n) for n = 1..1209 (full sequence)

Cf. A020473, A037264, A038034.

Subsequence of A214959.

a037268 n = a037268_list !! (n-1)
a037268_list = filter ((== 1) . a168046) $
                      takeWhile (<= 999999999) a214959_list
-- Reinhard Zumkeller, Aug 02 2012

A037268 := proc(n) option remember: local d,k: if(n=1)then return 1: fi: for k from procname(n-1)+1 do d:=convert(k,base,10): if(not member(0,d) and add(1/d[j],j=1..nops(d))=1)then return k: fi: od: end: seq(A037268(n),n=1..50); # Nathaniel Johnston, May 28 2011

Select[Range[10000],Total[1/(IntegerDigits[#]/.(0->1))]==1&] (* Harvey P. Dale, Jul 23 2025 *)

lista(nn) = {for (n=1, nn, d = digits(n); if (vecmin(d) && (sum(k=1, #d, 1/d[k])==1), print1(n, ", ")););} \\ Michel Marcus, Jul 06 2015

from fractions import Fraction
def ok(n):
  sn = str(n)
  return False if '0' in sn else sum(Fraction(1, int(d)) for d in sn) == 1
def aupto(limit): return [m for m in range(1, limit+1) if ok(m)]
print(aupto(8824)) # Michael S. Branicky, Jan 22 2021

More terms from Christian G. Bower, Jun 15 1998

Two missing terms inserted by Nathaniel Johnston, May 28 2011

A092666 a(n) = number of Egyptian fractions 1 = 1/x_1 + ... + 1/x_k (for any k), with 0 < x_1 <= ... <= x_k = n.

Original entry on oeis.org

1, 1, 1, 2, 1, 7, 1, 10, 10, 26, 1, 107, 1, 83, 375, 384, 1, 1418, 1, 4781, 7812, 1529, 1, 33665, 9789, 4276, 27787, 168107, 1, 584667, 1, 586340, 1177696, 52334, 5285597, 14746041, 1, 218959, 13092673, 84854683, 1, 279357910, 1, 491060793, 2001103921

Offset: 1

Max Alekseyev, Mar 02 2004

nonn

			a(4) = 2 since there are two fractions 1=1/2+1/4+1/4 and 1=1/4+1/4+1/4+1/4.

Toshitaka Suzuki, Table of n, a(n) for n = 1..390
Index entries for sequences related to Egyptian fractions

Cf. A020473, A092667, A092669, A002966, A000058, A259633.

a(n) = A020473(n) - A020473(n-1).

a(n) = 1 if n is prime.

Edited by Max Alekseyev, May 05 2010

A092670 a(n) = number of Egyptian fractions 1 = 1/x_1 + ... + 1/x_k (for any k), 0

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 6, 6, 6, 11, 11, 22, 22, 22, 22, 41, 41, 41, 41, 114, 114, 200, 200, 200, 363, 363, 566, 852, 852, 852, 852, 1655, 1655, 3054, 3054, 3054, 5777, 5777, 5777, 10647, 10647, 10647, 10647, 19436, 19436, 33373, 48360, 90441

Offset: 1

Max Alekseyev, Mar 02 2004

nonn

			a(6)=2 since there are two fractions 1=1/1 and 1=1/2+1/3+1/6.

Toshitaka Suzuki, Table of n, a(n) for n = 1..610
Index entries for sequences related to Egyptian fractions

Cf. A020473, A038034, A092669, A006585.

a(n)=Sum(A092669(i), i=1..n)

More terms from T. Suzuki (suzuki(AT)scio.co.jp), Nov 24 2006

A208480 Number of representations of n as m_1/1 + m_2/2 + ... + m_n/n where m_i are nonnegative integers.

Original entry on oeis.org

1, 3, 10, 55, 196, 2730, 10032, 108999, 973258, 20780331, 79309308, 2614200602, 10073335754, 288845706742, 11805287917646, 254331289285523, 997748503737330, 45462847634946017, 179126790937098750, 9868487079517742715, 566191988798860895880, 24134174584419467066865

Offset: 1

Brendan McKay and Max Alekseyev, May 07 2012

nonn

Number of unordered partitions of n into reciprocals of positive integers <= n.

N. K. Sinha et al., A simple looking problem in partitions that became increasingly complex, MathOverflow, 2012.

Cf. A020473, A269926, A378842.

More terms from Jinyuan Wang, Dec 12 2024

A119983 Number of ways to partition 1 into reduced fractions i/j with j <= n.

Original entry on oeis.org

1, 2, 4, 7, 13, 22, 36, 59, 107, 189, 244, 494, 594, 1063, 3276, 5508, 5804, 12427, 12916, 42411, 131773, 167588, 168842, 428013, 839368, 1015502, 1968162, 5787287, 5791851, 15163759, 15170600, 28838713, 75983560, 82753548, 486356263, 1158442727, 1158464363

Offset: 1

Franklin T. Adams-Watters, Aug 01 2006

nonn

The reduced fractions are the Farey fractions of order n (A005728). - Robert G. Wilson v, Aug 30 2010

			a(3) = 4; 1 = 1/1 = 1/2 + 1/2 = 2/3 + 1/3 = 1/3 + 1/3 + 1/3.

Cf. A000041, A020473, A115855 (one less), A115856.

Cf. A154886, A154888. - Reinhard Zumkeller, Jan 17 2009

Farey[n_] := Union@ Flatten@ Table[a/b, {b, n}, {a, b}]; f[n_] := Length@ IntegerPartitions[1, All, Farey@ n]; Array[f, 27] (* Robert G. Wilson v, Aug 30 2010 *)

For p prime, a(p) = a(p-1) + P(p) - 1, where P is the partition function (A000041).

Definition corrected by Reinhard Zumkeller, Jan 17 2009
a(21)-a(27) from Robert G. Wilson v, Aug 30 2010
More terms from Jinyuan Wang, Dec 12 2024

A212606 Number of distinct sums <= 1 of reciprocals of positive integers <= n.

Original entry on oeis.org

1, 2, 3, 6, 10, 26, 34, 103, 175, 393, 599, 2015, 2551, 8681, 14254, 19620, 34700, 129557, 161272, 595304, 695175, 1094164, 1903859, 7654850, 9413484, 29625309

Offset: 0

Max Alekseyev, May 22 2012

			a(3) = 6 counts numbers { 0, 1/3, 1/2, 2/3, 5/6, 1 }, each of which is can be represented as the sum of reciprocals 1/1, 1/2, and 1/3.

For distinct sums of distinct reciprocals, see A212607.

Cf. A020473, A092669, A092671, A208480

a(24)-a(25) from Dexter Senft, Feb 07 2019

A212607 Number of distinct sums <= 1 of distinct reciprocals of integers <= n.

Original entry on oeis.org

1, 2, 3, 5, 8, 14, 21, 38, 70, 129, 238, 440, 504, 949, 1790, 2301, 4363, 8272, 12408, 23604, 26675, 45724, 87781, 168549, 181989, 351076, 677339, 1306894, 1399054, 2709128, 2795144, 5423805, 10525050

Offset: 0

Max Alekseyev, May 22 2012

nonn

			a(3) = 5 counts numbers { 0, 1/3, 1/2, 5/6, 1 }, each of which is can be represented as the sum of distinct reciprocals 1/1, 1/2, and 1/3.

For possibly non-distinct reciprocals, see A212606.

Cf. A020473, A092669, A092671, A208480.

s:= proc(n) option remember;
      `if`(n=0, {0}, map(x-> `if`(n-1 nops(s(n)):
seq(a(n), n=0..20);  # Alois P. Heinz, May 23 2012

s[_] := s[n] = If[n == 0, {0}, If[n-1 < n*#, #, {#, # + 1/n}]& /@ s[n-1] // Flatten];
a[n_] := Length[s[n]];
Table[Print[n, " ", a[n]]; a[n], {n, 0, 32}] (* Jean-François Alcover, May 13 2019, after Alois P. Heinz *)

a(27)-a(32) from Alois P. Heinz, May 23 2012

A212657 Number of subsets of {1,2,...,n} with the sum of reciprocals <= 1.

Original entry on oeis.org

1, 2, 3, 5, 8, 14, 26, 46, 83, 151, 276, 503, 921, 1689, 3113, 5730, 10549, 19441, 35868, 66209, 122316, 226157, 418373, 774394, 1434130, 2657246, 4925837, 9135444, 16949660, 31460444, 58415377, 108502932, 201604081, 374708242, 696650259, 1295562800, 2410001851, 4484208954, 8345621293

Offset: 0

Max Alekseyev, May 23 2012

nonn

The number of distinct sums of reciprocals is given by A212607.

a(n) grows as 2^(b*n) with b=0.911... (Tikhomirov et al. 2017).

Robert Gerbicz, Table of n, a(n) for n = 0..95
M. Tikhomirov et al., Sets of unit fractions with sum <=1, MathOverflow, 2017.

Cf. A212658 (reciprocals can appear multiple times).

Cf. A212606, A212607, A020473, A092669, A092671, A208480.

{ A212657(n) = my(L=lcm(vector(n,i,i))); polcoeff( prod(i=1,n, 1+x^(L/i)+O(x^(L+1)) )/(1-x), L); }

a(32)-a(35) from Alois P. Heinz, May 23 2012

a(36)-a(95) from Robert Gerbicz, May 23 2012

A378270 Number of partitions of 1 into {1/1^2, 1/2^2, 1/3^2, ..., 1/n^2}.

Original entry on oeis.org

1, 2, 3, 7, 8, 58, 59, 259, 664, 3427, 3428, 73351, 73352, 298785, 7060868, 43070304, 43070305, 901194373, 901194374, 32808600352, 1204438945226, 2459587779124, 2459587779125, 96010353352980

Offset: 1

Ilya Gutkovskiy, Nov 21 2024

From David A. Corneth, Nov 24 2024: (Start)
Primes n/2 < p <= n occur in exactly one solution namely (p^2) * (1/p^2).
Proof: If the numerator k of k/p^2 is less than p^2 then p divides the denominator of the sum of the Egyptian fractions as p divides no other number <= n. But the goal is have sum 1 i.e. denominator 1 so p cannot be a divisor of the denominator. Consequently this can be reduced to "Number of partitions of 1 into {1/1^2, 1/2^2, ..., 1/(n/2)^2, ..., 1/n^2}" plus number of primes n/2 < p <= n. The denominators for the first part can be cleared turning this into a partitioning problem of positive integers. (End)

			a(4) = 7 because we have 16 * (1/16) = 12 * (1/16) + 1/4 = 8 * (1/16) + 2 * (1/4) = 4 * (1/16) + 3 * (1/4) = 9 * (1/9) = 4 * (1/4) = 1.
From _David A. Corneth_, Nov 24 2024: (Start)
To find a(12) we can rewrite the problem as "Number of partitions of 1 into {1/1^2, 1/2^2, 1/3^2, 1/4^2, 1/5^2, 1/6^2, 1/8^2, 1/9^2, 1/10^2, 1/12^2} + |{7, 11}|". The lcm of (1^2, 2^2, 3^2, 4^2, 5^2, 6^2, 8^2, 9^2, 10^2, 12^2) is 129600. So this comes a partition problem of (number of partitions of 129600 into parts 129600, 32400, 14400, 8100, 5184, 3600, 2025, 1600, 1296, 900) + |{7, 11}|. (End)

Index entries for sequences related to Egyptian fractions

Cf. A000290, A020473, A038034, A348625, A378271.

a(p) = a(p-1) + 1 for prime p. - David A. Corneth, Nov 22 2024

a(12)-a(21) from David A. Corneth, Nov 22 2024

a(22)-a(24) from Jinyuan Wang, Dec 11 2024

cp's OEIS Frontend

A038034 Number of compositions (ordered partitions) of 1 into {1/1, 1/2, 1/3, ..., 1/n}.

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A037268 Sum of reciprocals of digits = 1.

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A092666 a(n) = number of Egyptian fractions 1 = 1/x_1 + ... + 1/x_k (for any k), with 0 < x_1 <= ... <= x_k = n.

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A092670 a(n) = number of Egyptian fractions 1 = 1/x_1 + ... + 1/x_k (for any k), 0

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A208480 Number of representations of n as m_1/1 + m_2/2 + ... + m_n/n where m_i are nonnegative integers.

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Extensions

A119983 Number of ways to partition 1 into reduced fractions i/j with j <= n.

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Programs

Mathematica

Formula

Extensions

A212606 Number of distinct sums <= 1 of reciprocals of positive integers <= n.

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Extensions

A212607 Number of distinct sums <= 1 of distinct reciprocals of integers <= n.

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Programs

Maple

Mathematica

Extensions

A212657 Number of subsets of {1,2,...,n} with the sum of reciprocals <= 1.